Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So the binding energy of an object is the amount of energy needed to move it an infinite distance away from another mass to essentially “escape” its gravitational field. But what happens if you give the object more energy than the binding energy. Will the object go “past infinity”?

share|cite|improve this question
It slows down until its kinetic energy equals its initial kinetic energy minus the energy needed to climb out of the gravity well. Then it just keeps going at that speed. – Mike Dunlavey Jun 25 '12 at 17:11
Look at velocity equation of an object thrown straight up from Earth surface – voix Jun 25 '12 at 17:30
up vote 2 down vote accepted

I'd just like to slightly refine John and Mike's answers. If the initial kinetic energy equals or exceeds the binding energy, then the particle speed asymptotically approaches a constant, never stopping and "turning around". Of course, in the very special case that the KE equals the binding energy, that constant speed is precisely zero.

Note that though the particle never stops moving, it's always, always a finite distance from the other mass. Not only can it not go "past" infinity, it can't ever get there!

share|cite|improve this answer

If the kinetic energy of the object is exactly the same as the binding energy, the object will slow as it leaves the gravity well and will come to a halt at infinity (i.e. far enough away for the gravity to be insignificant).

If the kinetic energy of the object is greater than the binding energy, the object will still slow as it leaves the gravity well, but it will settle down to a constant speed at infinity.

So the excess energy doesn't mean the object will go "past infinity", it just means that it's velocity settles to a constant value instead of becoming zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.